MCQ: Height & Distance - 2 - SSC CGL MCQ

# MCQ: Height & Distance - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Height & Distance - 2

MCQ: Height & Distance - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Height & Distance - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Height & Distance - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Height & Distance - 2 below.
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MCQ: Height & Distance - 2 - Question 1

### A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 45° to 60°, how soon after this will the car reach the observation tower?

Detailed Solution for MCQ: Height & Distance - 2 - Question 1

Let AB be the tower and C and D be the two positions of the car.
Then, from figure
AB/AC = tan 60 = √3 => AB = √3AC
AB = AC + CD
CD = AB - AC = √3AC - AC = AC (√3 - 1)
CD = AC (√3 - 1) => 10 min
AC => ?
AC/(AC(√3 - 1)) x 10 =? = 10/(√3 - 1) = 13.66 = 13 min 20 sec(approx)

MCQ: Height & Distance - 2 - Question 2

### Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Detailed Solution for MCQ: Height & Distance - 2 - Question 2

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, ACB = 30° and ADB = 45°.

AC = AB x 3 = 1003 m.

CD = (AC + AD)= (1003 + 100) m
= 100(3 + 1)
= (100 x 2.73) m
= 273 m.

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MCQ: Height & Distance - 2 - Question 3

### From The highest point of a 10 m high building, the edge of rise of the of the highest point of a tower is 60° and the despondency's edge of its foot is 45°, Find The tower's stature. (take√3 = 1.732)

Detailed Solution for MCQ: Height & Distance - 2 - Question 3

Let AB be the building and CD be the tower.
Draw BE perpendicular to CD.
At that point CE = AB = 10m, ∠EBD = 60° and ∠ACB = ∠ CBE = 45°
AC/AB = cot45° = 1 = >AC/10 = 1 => AC = 10m.
From △ EBD, we have
DE/BE = tan 60° = √3 => DE/AC = √3
=> DE/10 = 1.732 => DE = 17.3
Height of the tower = CD = CE + DE = (10 + 17.32) = 27.3 m.

MCQ: Height & Distance - 2 - Question 4

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

Detailed Solution for MCQ: Height & Distance - 2 - Question 4

One of AB, AD and CD must have given.

MCQ: Height & Distance - 2 - Question 5

The horizontal distance between two towers is 90 m. The angular depression of the top of the first as seen from the top of the second which is 180 m high is 450. Then the height of the first is

Detailed Solution for MCQ: Height & Distance - 2 - Question 5

=> (180 - h)/90 = Tan(45)
=> h = 90 m

MCQ: Height & Distance - 2 - Question 6

The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree, is:

Detailed Solution for MCQ: Height & Distance - 2 - Question 6

Let AB be the tree and AC be its shadow.

Let ACB = θ.
Then, AC = 3          cot θ = 3AB

θ = 30º.

MCQ: Height & Distance - 2 - Question 7

A man is watching form the top of the tower a boat speeding away from the tower. The boat makes the angle of depression of 60° with the man's eye when at a distance of 75 meters from the tower. After 10 seconds the angle of depression becomes 45°. What is the approximate speed of the boat, assuming that it is running in still water?

Detailed Solution for MCQ: Height & Distance - 2 - Question 7

Let AB be the tower and C and D be the positions of the boat.
Distance travelled by boat = CD
From the figure 75tan(60) = (75 + CD)tan(45)
=> 75√3 = 75 + CD
=> CD = 55 m
Speed = distance/time = 55/10
= 5.5 m/sec = 19.8 kmph

MCQ: Height & Distance - 2 - Question 8

From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Detailed Solution for MCQ: Height & Distance - 2 - Question 8

Let AB be the tower.

Then, APB = 30º and AB = 100 m.

AP= (AB x √3) m
= 100√3 m
= (100 x 1.73) m
= 173 m.

MCQ: Height & Distance - 2 - Question 9

An observer 1.4 m tall is 10√3 away from a tower. The angle of elevation from his eye to the top of the tower is 60°. The heights of the tower is

Detailed Solution for MCQ: Height & Distance - 2 - Question 9

Let AB be the observer and CD be the tower.
Then, CE = AB = 1.4 m,
BE = AC = 10v3 m.
DE/BE = Tan (30) = 1/√3
DE = 10√3/√3 = 10
CD = CE + DE = 1.4 + 10 = 11.4 m

MCQ: Height & Distance - 2 - Question 10

The Top of a 25 meter high tower makes an angle of elevation of 450 with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pole.

Detailed Solution for MCQ: Height & Distance - 2 - Question 10

Let AB be the tower and CD be the electric pole.
From the figure CA = DE
=> 25/(Tan(45)) = (25-h)/(Tan(30))
=> 25  Tan(30) = 25 - h
=> h = 25 - 25Tan(30)
= 25(1 - Tan(30))
= 25((√3 - 1)/√3)

MCQ: Height & Distance - 2 - Question 11

From a point P on a level ground, the angle of elevation of the top tower is 60°. If the tower is 180 m high, the distance of point P from the foot of the tower is

Detailed Solution for MCQ: Height & Distance - 2 - Question 11

From ∠APB = 60° and AB = 180 m.
AB/AP = tan 60° =√3
AP = AB/√3 = 180/√3 = 60√3

MCQ: Height & Distance - 2 - Question 12

The heights of two towers are 90 meters and 45 meters. The line joining their tops make an angle 450 with the horizontal then the distance between the two towers is

Detailed Solution for MCQ: Height & Distance - 2 - Question 12

Let the distance between the towers be X
From the right angled triangle CFD
Tan(45) =  (90 - 45)/X
=> x = 45 meters

MCQ: Height & Distance - 2 - Question 13

The angles of elevation of the tops of two vertical towers as seen from the middle point of the lines joining the foot of the towers are 45° & 60°.The ratio of the height of the towers is

Detailed Solution for MCQ: Height & Distance - 2 - Question 13

Tan(60) = h1/AB
=> h1 = √3AB
Tan(45) = h1/BC
=> h2 = BC
h1/ h2 = √3/1
=> h1 : h2 = √3 : 1

MCQ: Height & Distance - 2 - Question 14

On the level ground, the angle of elevation of the top of a tower is 30°.on moving 20 meters nearer, the angle of elevation is 60°. Then the height of the tower is

Detailed Solution for MCQ: Height & Distance - 2 - Question 14

Let h be the height of tower
From figure.
20 = h (  cot30 - cot60)
20 = h (√3 - 1/√3)
=> 20√3 = h (3 - 1)
=> h = 10√3.

MCQ: Height & Distance - 2 - Question 15

The angle of elevation of a tower at a point 90 m from it is cot-1(4/5). Then the height of the tower is

Detailed Solution for MCQ: Height & Distance - 2 - Question 15

Let cot-1(4/5) = x
=> cot x = 4/5
=> tan(x) = 5/4
From the right angled triangle
Tan(x) = h/90
=> h = 5/4 * 90 =112.5 m

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests